- Organic Chemistry
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- Amines and other nitrogen compounds
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- Chemistry in daily life
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# Chemical and Ionic Equilibrium

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**Asked By: ROHIT SINGH**7 Month ago

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**Asked By: ROHIT SINGH**7 Month ago

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A weak acid, HA after treatment with 20 ml of 0.1 M strong base BOH has a pH of 5. At the end poin the volume of same base required is 27 ml. What is the K_{a }of acid?

**Asked By: MRUNMAYEE R GAIKWAD**1 year ago

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specific heat of a metal is 0.031cal/degree.gram its equivalent weight is 103.6.calculate its exact atomic weight of metal

**Asked By: CHANDRESH DHAR DUBEY**1 year ago

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can pH >7 and pOH<7?

**Asked By: NEHA VISHWAS GURAV**1 year ago

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**Degree when a small amount of HCl is added to a buffer solution of acetic acid and sodium acetate**

**(i) pH increases**

**(ii) [H ^{+}] decreases**

**(iii) Dissociation of acetic acid decreases**

**Asked By: SWATI KAPOOR**1 year ago

**Solved By: SARIKA**

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**If ionic product of water is K _{w} = 10^{-6} at 4°C, then a solution with pH = 7.5 at 4°C will**

**(i) turns blue litmus red **

**(ii) turns red litmus blue**

**(ii) Be neutral t****o litmus**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

question is not appropriate because for given K_{w} , [H^{+}] will be 10^{-3 }and max. pH will be 6 and the question is asking about a solution of pH 7.5 which is out of limit

but if it is considered as a correct question then answer will be (ii)

**N _{2} + O_{2} = 2NO . Equilibrium constant K_{c} = 2 . Degree of dissociation is**

**(i) 1/1-****√2 (ii) ****1/1+****√2 (iii) 2****/1-****√2**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

** N _{2} + O_{2} = 2NO **

1 1 0

1-x 1-x 2x

so by formula K_{c} = (2x)^{2} / (1-x)(1-x)

now solve for x

**In the equilibrium SO _{2}Cl_{2} = SO_{2} + Cl_{2} at 2000K and 10 atm pressure , % Cl_{2} = % SO_{2} = 40 (by volume) . Then what is the value of K_{p}?**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

**SO _{2}Cl_{2} = SO_{2} + Cl_{2}**

_{100 0 0}

_{20 40 40}

_{2 atm 4 atm 4 atm}

**now calculate K _{p} **

**0.2 mole of N _{2} and 0.6 mole of H_{2} react to give NH_{3} and 40 % of reactant mixture is decreased , according to the equation, **

**N _{2} (g) + 3H_{2} (g) = 2NH_{3} (g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

**N _{2} (g) + 3H_{2} (g) = 2NH_{3} (g)**

**0.2 0.6 0**

**0.2-x 0.6-3x 2x**

according to the given condition 4x = 0.8*40/100 = 0.32

so x = 0.08

now solve

1 mole of A , 1.5 mole of B and 2 moles of C are taken in a vessel volume one litre. At equilibrium concentration of C is 0.5 mole/L. Equilibrium constant for the reaction

**A (g) + B (g) = C (g)**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

**A (g) + B (g) = C (g)**

** 1 1.5 2**

**1-x 1.5-x 2+x**

**according to the given condition 2+x = 0.5 so x = -1.5**

**now calculate K _{c}**

The K_{sp} for a sparingly soluble Ag_{2}CrO_{4} is 4 x 10^{-12}. The molar solubility of the salt is

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

Ag_{2}CrO_{4 }-------------> 2Ag + CrO_{4}

s 0 0

2s s

so K_{sp }= (2s)^{2} s = 4s^{3}

now solve

A 0.1N solution of sodium bicarbonate has a pH value of

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

Please submit complete question, K_{a} should be given with percentage of ionisation

pH of 0.1 M NH_{4}Cl solution is

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

For calculating pH of a salt, only concentration is not sufficient

**pH of 10 ^{2} M HCl is**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

This is a strong acid so consider 100 percent ionisation

we get [H^{+}] = 10^{2}

it is more than 1 so pH = 0

**100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

mili gm eq of NaOH = NV = 10

mili gm eq of HCl = 20

total mili gm eq = 20-10 = 10 of HCl

so N = 10/200 = 0.05

so pH = log[1/0.05]

1 c.c of 0.1 N HCl is added to 1 litre solution of sodium chloride. The pH of the resulting solution will be

**Asked By: SWATI KAPOOR**1 year ago

**Solved By: SWATI KAPOOR**

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When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. The solution is found to have a hydroxide ion concentration of 1.34 x 10^{-3} . The dissociation constant of ammonia is

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

NH_{3} + H_{2}O---------------------> NH_{4}^{+} + OH^{-}

0.1 0 0

0.1(approx) 1.34*10^{-3} 1.34*10^{-3}

now calculate dissociation constant

**Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. K _{a} of acetic acid is 1.8 x 10^{-5}**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

CH_{3}COOH ----------------> CH_{3}COO^{-} + H^{+}

0.01 0 0.1

0.01-x x 0.1+x

0.1+x =0.1 for x is negligibly small

so K_{a }= x*0.1/(0.01-x)

now solve

The dissociation constant of a weak acid HA and weak base BOH are 2 x 10^{-5} and 5 x 10^{-6} respectively. The equilibrium constant for the neutralization reaction of the two is (ignore hydrolysis of resuting salt)

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

HA + BOH ------------------> B^{+} + A^{-} + H_{2}O

K = [B^{+}][A^{-}][H_{2}O]/[HA][BOH]

= [B^{+}][OH^{-}]/[BOH] * [H^{+}][A^{-}]/[HA] * [H_{2}O]/[H^{+}][OH^{-}] = K_{b} * K_{a} / K_{w}

now solve

**Which of the following increasing order of pH of 0.1 M solution of the component (a) HCOONH _{4} (b) CH_{3}COOHNH_{4 }(c) CH_{3}COONa (d) NH_{4}Cl is correct**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

**(d) NH _{4}Cl < **

**(a) HCOONH**

_{4 }< (b)**CH**

_{3}COONH_{4 }<_{ }**(c) CH**

_{3}COONa
Which of the following molar ratio of NH_{3} and HCl in aqueous solution will constitute a buffer ? (a) 1 : 2 (c) 1 : 1 (b) 1 : 3 (d) 2 : 1

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

in (d) after reaction no. of moles of NH_{3 }will be 1 and 1 mole NH_{4}Cl will be formed so it will be a basic buffer.

**In order to prepare a buffer of pH 8.26, the amount of (NH _{4} )_{2}SO_{4} required to be mixed with 1L of 0.1M NH_{3} (pK_{b} of NH_{3 }= 4.74) is**

**Asked By: SWATI KAPOOR**1 year ago

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**Solution by manish sir**

**NH _{3} + H_{2}O ** ------------> NH

_{4}

^{+}+ OH

^{-}

so K_{b} = [NH_{4}^{+}] [OH^{-}]/[NH_{3}][H_{2}O]

so K_{b} = [NH_{4}^{+}]/[NH_{3}][H^{+}] * [H^{+}][OH^{-}]/[H_{2}O]

so K_{b} = [NH_{4}^{+}]/[NH_{3}][H^{+}] * Kw

now take log and solve. concentration of **(NH _{4} )_{2}SO_{4 will be double of }**[NH

_{4}

^{+}]

if you find any problem on solving this then reply

1.0 ml of dilute solution of NaOH is added to 100 ml of a buffer of pH 4. The pH of the resulting solution

(a) becomes 7.0

(b) becomes 9.0

(c) becomes 3.0

(d) Remains practically unchanged

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

it is strong acidic buffer so answer is (d)

**10 ml of M/200 H _{2}SO_{4} is mixed with 40 ml of M/200 H_{2}SO_{4}. The ph of the resulting solution is**

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

M of H_{2}SO_{4} = 1/200

so [H^{+}] = 2*1/200 = 1/100 = 10-2

now calculate pH

The pH of solution at 25°C which has twice as many hydroxide ion as in pure water at 25°C , will be

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

no of hydroxide ion in pure water = 10^{-7}

so according to the given condition no. of hydroxide ions in solution = 2*10^{-7}

so pOH = -log2*10^{-7}

so pH = 14 - pOH

The ph of 0.1 M HCN is 5.2. What is the K_{a}?

**Asked By: SWATI KAPOOR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

HCN -------------------> H^{+} + CN^{-}

0.1 0 0

0.1-x x x

pH = 5.2

so -log[H^{+}] = 5.2

solve for [H^{+}], it will give x

then calculate K_{a}

what is the ph of the solution formed by mixing 20 ml of 0.05 M H_{2}SO_{4} with 5 ml of 0.45 M NaOH at 298 k ?

**Asked By: SAMREEN KOUR**1 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

mili eq of acid = VN = 20*0.1 = 2

mili eq of base = VN = 5*0.45 = 2.25

total milieq = 2.25-2 = 0.25 of base in 25 ml

so N = 0.25/25 = 10-2

so pOH = 2 so pH = 12

1-In the following hypothetical reaction A + 3B --------------

**Asked By: SHWETA BHARDWAJ**2 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

A + 3B ---------> 2C + D

1 1/2 0 0

1-x 1/2-3x 2x x

according to the given condition 1/2 - 3x = 2x

so 1/10 = x

so 3x = 3/10

so % = 0.3*100/0.5 = 60

when 20 g of caco3 were put into 10 litre flask and heated to 800 degree cesius, 30% of caco3 remained unreacted at equilibrium. kp for decomposition of caco3 will be 1.231 ATM HOWWWW............PLS SOLVE THIS ....PLS PLS..........

**Asked By: PRATIBHA**2 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

CaCO_{3} ----------> CaO + CO_{2 }(g)

0.2 mole 0 0

0.06 0.14 0.14

so K_{p}= p only CO_{2} is gas so its pressure is considered

now p = 0.14*0.0821*1073/10

now solve

Hello,

sir can someone explain to me what is ionic equilibria. And how to calculate the ph and -log c. And so on. I am really confused. Thank you.

**Asked By: MARWA**2 year ago

**Solved By: SARIKA SHARMA**

**read solutions ( 2 ) | submit your answer**

QUES) AMMONIA UNDERGOES SELF DISSOCIATION ACCORDING TO D Rxn

2NH_{3} (l) ------> NH_{4}+ (am) + NH_{2}^{-} (am)

WHERE am , STANDS FOR AMMONIATED . WHEN 1 MOL OF NH4Cl IS DISSOLVED IN 1 kg OF LIQUID AMMONIA , THE B.P. AT 760 TORR IS OBSV. AT -32.7^{0} C ( NORMAL BOILING B.P OF NH3 (l) IS -33.4^{0} C )

WHAT CONCLUSn ARE REACHED ABOUT THE NATURE OF SOLn??

( ANS NH4Cl IS COMPLETELY DISSOCIATED IN NH3)

**Asked By: SARIKA SHARMA**2 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

boiling point elevation = (-32.7) - (-33.4) = 0.7

now according to the given data

boiling point elevation = 1000*K(ammonia)*1/1 = 1000*0.00035 = 0.35

so i = 0.7/0.35 = 2

so 100 % dissociation

SIR , I WANT TO KNOW THAT IN THE EQUILM QUES WHY YOU TAKE MOL FRACT FOR CH3COCH3=1-1/6-1/6.WHY YOU SUB FROM 1.

**Asked By: SARIKA**2 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

If we multiply mole fraction to pressure, we get the pressure exerted by the component so it is a good way to get answer. and 1 represents net sum of mole fractions of all the components. here first and second 1/6 stands for mole fraction of ethane and carbon monooxide.

Q CH3COCH3 (g) eqm CH3CH3 +CO

initial pressure of CH3COCH is 300 mm when eqm is setup mole fraction of CO(g) = 1/6 hence total pressure will be =?

ANS 360mm of hg .SOL =?

**Asked By: SARIKA**2 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

At equilibrium mole fraction of CO = 1/6, that of CH_{3}CH_{3 }= 1/6

so mole fraction of CH_{3}COCH_{3 } = 1- 1/6 - 1/6 = 2/3

let the total pressure at equilibrium = p

so pressure of CH_{3}CH_{3} and CO will be p/6 and p/6

and that of CH_{3}COCH_{3 }= 2p/3

now consider the following dissociation data

CH_{3}COCH_{3} ----> CH_{3}CH_{3} + CO

300 mm 0 0 at starting

300-x mm x x at equilibrium

here x = p/6 (1) and 300-x = 2p/3 (2) add these equations and get the value of p

Sir can you please post a topic on ionic equilibrium ( especially for numerical solving).

**Asked By: AMIT DAS**2 year ago

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**Solution by manish sir**

YES

very soon

what is difference between dissociatio and ionisation?

**Asked By: ANUJ KUMAR YADAV**3 year ago

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**Answer Strategies and trick**(it will help you to solve it by yourself)

Dissociation is the reaction in which two or more than two product molecules are formed.

ex. PCl_{5} ---> PCl_{3} + Cl_{2}

whereas ionisation is the reaction in which ions are formed

ex. MgCl_{2} ----> Mg^{++} + 2 Cl^{-}